Problem 30 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$y=\frac{1}{x} \sin ^{-5} x-\frac{x}{3} \cos ^{3} x$$

Step-by-Step Solution

Answer

To be completed after solution steps.

Step 1: Identify the Function Components

Let's begin by identifying the two main components in the given function. The function is given by:\[ y = \frac{1}{x} \sin^{-5}(x) - \frac{x}{3} \cos^3(x) \]This function consists of two separate terms, which we'll handle individually: \( u = \frac{1}{x} \sin^{-5}(x) \) and \( v = \frac{x}{3} \cos^3(x) \).

Step 2: Identify the relevant trigonometric identities

Based on the given expression or equation, identify which trigonometric identities (Pythagorean, double-angle, sum/difference, etc.) are applicable.

Step 3: Apply the identities and simplify

Apply the identified identities to transform the expression. Simplify step by step, combining like terms and reducing fractions where possible.

Step 4: Solve or evaluate

If solving an equation, isolate the trigonometric function and find the angle(s). If evaluating, compute the final numerical value.

Step 5: State the result

Express the final answer, including all solutions in the required domain if solving an equation.

Step 6: Conclude with the answer

To be completed after solution steps.

Key Concepts

Trigonometric Functions

Trigonometric functions describe the relationships of the angles and sides in right triangles. In calculus, trigonometric functions like sine (sin), cosine (cos), and tangent (tan) play a crucial role. They provide real-life applications in oscillatory functions, such as sound waves.
In the exercise, we see the use of \(\sin^{-5}(x)\) and \(\cos^3(x)\). The notation \(\sin^{-5}(x)\) implies the sine function raised to the -5th power. It is equivalent to \((\sin(x))^{-5}\), meaning you take the sine of \(x\) and raise it to the power of -5. Similarly, \(\cos^3(x)\) means \((\cos(x))^3\).
Trigonometric functions are periodic, which means they repeat their values in regular intervals. This feature is particularly essential when finding derivatives, as these derivatives also display periodic behavior. Understanding these functions is vital to solving calculus problems efficiently.

Power Rule

The power rule is a basic differentiation rule used for finding the derivative of a function of the form \(f(x) = x^n\). The rule states:

• If \(n\) is any real number, then the derivative of \(f(x) = x^n\) is \(f'(x) = nx^{n-1}\).

For instance, if you have \(x^3\), applying the power rule gives \(3x^2\).
The power rule is particularly useful in this exercise when dealing with terms such as \(\sin^{-5}(x)\) in \(\frac{1}{x} \sin^{-5}(x)\). Here, the 'power' is -5, which means you'd bring down -5 as a multiplier of the sine term and then subtract one from the power to differentiate it.
Using the power rule simplifies processes by reducing complicated powers into simpler, more manageable terms. This makes solving derivatives much more straightforward.

Product Rule

When you have a function that is a product of two or more functions, we use the product rule to find derivatives. The product rule states:

• For functions \( u(x) \) and \( v(x) \), the derivative of their product is \((uv)' = u'v + uv'\).

We apply the product rule to each term in the given function that involves a product of functions.
For \( u = \frac{1}{x} \sin^{-5}(x) \), let \(a = \frac{1}{x}\) and \(b = \sin^{-5}(x)\). Use the product rule:

• The derivative \( u' = a'b + ab' \).

Similarly, apply it to \( v = \frac{x}{3} \cos^3(x) \). Each part becomes manageable using this rule, making complex derivatives understandable. The product rule is crucial for tackling problems where components are multiplied.

Chain Rule

The chain rule is used when dealing with composite functions, helping to find the derivative of one function that is within another. Its essence is:

• If you have a composite function \( y = f(g(x)) \), then the derivative is \( y' = f'(g(x)) \cdot g'(x) \).

This is important for problems where trigonometric functions are raised to a power, like our case with \( \cos^3(x) \) and \( \sin^{-5}(x) \). You treat these as composite functions because \( \cos(x) \) or \( \sin(x) \) is within another operation, raising it to a power.
Taking the derivative of \( \cos^3(x) \), for example, you'd first treat it as \((g(x))^3 \) where \( g(x) = \cos(x) \), differentiate \(g\), and multiply by the outside function's derivative. This breaks complex derivatives into simple steps, making them much easier to compute.